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Demand Equation The price p (in dollars) and the quantity

x sold of a certain product obey the demand equation

p=-110x+1500≤x≤1500

(a) Express the revenue R as a function of x.

(b) What is the revenue if 100 units are sold?

(c) What quantity x maximizes revenue? What is the

maximum revenue?

(d) What price should the company charge to maximize

revenue?

Short Answer

Expert verified

(a) R=-110x2+150x

(b) R=14000

(c) x=750&R=56250

(d) To maximize revenue, pricerole="math" localid="1647360230117" =$75

Step by step solution

01

Part (a) Step 1. Given Information

p=-110x+150

02

Part (a) Step 2. Calculation 

R=xp=x(-110x+150)=(-110x2+150x)

03

Part (b) Step 1. Given Information

p=-110x+150

04

Part (b) Step 2. Calculation

For x=100,

R=-110×10000+150×100

=14000

05

Part (c) Step 1. Given Information

p=-110x+150

06

Part (c) Step 2. Calculation

To maximize revenue,

R'(x)=-15x+150, which is equal to zero

or,-15x+150=0or,x=750

and maximum revenue isR=56250

07

Part (d) Step 1. Calculation

p=-110x+150

08

Part (d) Step 2. Calculation

To maximize revenue x=750

Then p=-110x+150

or,p=-110×750+150=75

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